Problem 50 - A Complete Resource Book in Chemistry for JEE Main

Question

Which one of the following is the correct set with reference to molecular formula, hybridization of central atom and shape of the molecule? (a) \(\mathrm{CO}_{2}, \mathrm{sp}^{2}\), bent (b) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{2}\), bent (c) \(\mathrm{BeCl}_{2}, \mathrm{sp}\), linear (d) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{sp}^{3}\), linear

Step-by-Step Solution

Answer

(c) \(\mathrm{BeCl}_{2}, \mathrm{sp},\) linear is the correct set.

Step 1: Analyzing CO2

The central atom in \(\mathrm{CO}_{2}\) is carbon. CO2 has a linear shape as the carbon atom forms double bonds with two oxygen atoms. The hybridization of carbon in this molecule is \(\mathrm{sp}\), not \(\mathrm{sp}^{2}\). Thus option (a) is incorrect.

Step 2: Analyzing H2O with sp2 Hybridization

For \(\mathrm{H}_{2}\mathrm{O}\), the central atom is oxygen. The molecular geometry of water is bent due to the two lone pairs causing a bond angle of approximately 104.5°. However, the correct hybridization of the oxygen atom is \(\mathrm{sp}^{3}\), not \(\mathrm{sp}^{2}\). Thus option (b) is incorrect.

Step 3: Analyzing BeCl2

In \(\mathrm{BeCl}_{2}\), the central atom is beryllium. Beryllium forms single bonds with two chlorine atoms, creating a linear arrangement with a bond angle of 180°. The hybridization of beryllium in \(\mathrm{BeCl}_{2}\) is \(\mathrm{sp}\). Thus option (c) is correct.

Step 4: Analyzing H2O with sp3 Hybridization

Returning to \(\mathrm{H}_{2}\mathrm{O}\), as previously noted, the hybridization of the oxygen atom indeed is \(\mathrm{sp}^{3}\), but the geometry is bent, not linear. Therefore, option (d) is incorrect.

Key Concepts

Hybridization

Hybridization is a key concept in understanding the geometric configuration of molecules. It's a way of mixing atomic orbitals into new hybrid orbitals suitable for pairing electrons to form chemical bonds. This results in the particular shape of the molecule. For example, in \(\mathrm{BeCl_2}\), beryllium undergoes \(\mathrm{sp}\) hybridization. This involves mixing one s orbital with one p orbital, resulting in two equivalent \(\mathrm{sp}\) hybrid orbitals. Each of these hybrid orbitals forms a sigma bond with chlorine, creating a linear molecular structure.
In water, \(\mathrm{H_2O}\), oxygen undergoes \(\mathrm{sp^3}\) hybridization. Here, one s orbital and three p orbitals mix to create four equivalent \(\mathrm{sp^3}\) hybrid orbitals. Two of these orbitals hold lone pairs, while the other two form bonds with hydrogen atoms. This arrangement makes the structure bent, due to lone pair repulsions.
Understanding hybridization is crucial to predict molecular shapes and bonding, which in turn affects both physical and chemical properties of compounds.

VSEPR Theory

VSEPR stands for Valence Shell Electron Pair Repulsion theory. This model helps predict the geometry of molecules based on the repulsion between electron pairs in the valence shell of the central atom. The idea is that electron pairs will arrange themselves to be as far apart as possible to minimize repulsion.
In \(\mathrm{CO_2}\), carbon acts as the central atom and creates a linear shape with a bond angle of 180°. This is because the double bonds repel each other equally, distributing themselves in a straight line. For \(\mathrm{BeCl_2}\), which also features a linear configuration, the beryllium and its surrounding chlorine atoms align in a straight line due to \(\mathrm{sp}\) hybridization.
In contrast, \(\mathrm{H_2O}\) is not linear but bent. The oxygen atom has two lone pairs of electrons which create greater repulsions than bonded pairs. Therefore, according to VSEPR theory, the molecular structure bends to decrease repulsion, with a bond angle of about 104.5°.
By applying VSEPR theory, students can predict whether a molecule will be linear, bent, or take another shape entirely.

Chemical Bonding

Chemical bonding involves the attractive force holding atoms together in compounds or molecules. It explains how molecules form and their resulting properties. In covalent bonding, atoms share electron pairs to fulfill the octet rule, aiming for stability by having a full valence shell.
In \(\mathrm{CO_2}\), carbon forms two double covalent bonds with oxygen atoms. Each double bond consists of one sigma and one pi bond, making the molecule stable and linear. Similarly, beryllium in \(\mathrm{BeCl_2}\) forms single bonds with chlorine atoms, resulting in a straightforward linear structure due to equal bond lengths and energies in \(\mathrm{sp}\) hybridization.
Water's structure in \(\mathrm{H_2O}\) is slightly different due to its bent shape. Oxygen shares electron pairs with hydrogen, which results in two sigma bonds. Meanwhile, the two unshared electron pairs contribute to the molecular shape and bond angle.
Understanding chemical bonding is crucial for determining molecule stability, reactivity, and interaction with other compounds. It helps students comprehend how different elements form vastly different structures and properties.